A 0.5 kg wooden block is placed on top of a 1.0 kg wooden block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.2. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

I found the kinetic friction of the two blocks, which is 2.94N. I found the static friction of the upper block, which is 1.715 N. I don't know what to do next. Do you add them together? When I tried it, I got less than 8.1 N, which is the answer in the back.

To find the maximum horizontal force that can be applied to the lower block without the upper block slipping, we'll need to consider the forces acting on both blocks.

Let's start by calculating the maximum static friction force between the two blocks. The maximum static friction force can be found by multiplying the coefficient of static friction (μs) by the normal force (N) between the two blocks. In this case, the normal force between the two blocks is equal to the weight of the upper block.

The weight of the upper block can be calculated by multiplying its mass (m) by the acceleration due to gravity (g). Thus, the normal force N is given by:

N = (mass of upper block) * (acceleration due to gravity)
= (1.0 kg) * (9.8 m/s^2)
= 9.8 N

Now, we can calculate the maximum static friction force (Fs_max) between the two blocks:

Fs_max = μs * N
= 0.35 * 9.8 N
= 3.43 N

Next, we need to take into account the kinetic friction between the lower block and the table. Since the lower block is what we are applying the force to, we need to consider the force required to overcome kinetic friction.

The kinetic friction force (Fk) can be calculated by multiplying the coefficient of kinetic friction (μk) by the normal force (N) acting on the lower block. Remember, in this case, the normal force on the lower block is equal to its weight.

The weight of the lower block can be calculated as:

Weight = (mass of lower block) * (acceleration due to gravity)
= (0.5 kg) * (9.8 m/s^2)
= 4.9 N

Now, let's calculate the kinetic friction force (Fk):

Fk = μk * N
= 0.2 * 4.9 N
= 0.98 N

To keep the upper block from slipping, the applied force (Fa) must overcome both the static friction force and the kinetic friction force. So, we need to find the combined force that can be applied without causing either block to slip.

To find the maximum horizontal force, we sum the maximum static friction force and the kinetic friction force:

F_max = Fs_max + Fk
= 3.43 N + 0.98 N
= 4.41 N

So, the maximum horizontal force that can be applied to the lower block without the upper block slipping is approximately 4.41 N.